Question 1

Time gap to infer the slower speed: In a 100 m race, Kamal defeats Bimal by 5 s. If Kamal’s speed is 18 km/h, what is Bimal’s speed (km/h)?

Accepted Answer

Introduction / Context: Given one runner's speed and the finish time difference, compute the slower runner's finish time and speed. Given Data / Assumptions: Kamal speed = 18 km/h = 5 m/s. Race = 100 m. Kamal wins by 5 s, so Bimal takes 5 s longer. Concept / Approach: First find Kamal's time; add 5 s for Bimal; then compute Bimal's speed = distance / time, convert to km/h. Step-by-Step Solution: tKamal = 100 / 5 = 20 stBimal = 20 + 5 = 25 svBimal = 100 / 25 = 4 m/s = 14.4 km/h Verification / Alternative check: Unit conversion: 4 m/s * 3.6 = 14.4 km/h. Why Other Options Are Wrong: 15.4/14.5/14 are rounding or setup errors; the exact is 14.4. Common Pitfalls: Misconverting m/s ↔ km/h or misapplying the 5 s gap. Final Answer: 14.4 km/h

Question 2

Repeat of start + time-gap computation: In a 100 m race, A runs at 5 km/h, gives B a start of 8 m, and still beats B by 8 s. Find B’s speed.

Accepted Answer

Introduction / Context: This is the same structure as a prior item; we recompute to confirm consistency. Given Data / Assumptions: A’s speed = 5 km/h = 1.3889 m/s; B runs 92 m; B finishes 8 s after A. Concept / Approach: Compute A’s finish time for 100 m, then add 8 s for B’s finish time over 92 m. Step-by-Step Solution: tA ≈ 72 s ⇒ tB = 80 svB = 92/80 = 1.15 m/s = 4.14 km/h Verification / Alternative check: Matches earlier derived value. Why Other Options Are Wrong: Numbers that do not match the precise timing. Common Pitfalls: Mixing metres and kilometres without consistent conversions. Final Answer: 4.14 Km/h

Question 3

Transferring race results between contestants: In a 1 km race, A beats B by 100 m and C by 200 m. By how many metres can B beat C in a 1350 m race?

Accepted Answer

Introduction / Context: We derive B:C speed ratio from A’s known advantages over B and C, then project that ratio to a new race distance to find B’s margin over C. Given Data / Assumptions: A beats B by 100 m → when A runs 1000 m, B runs 900 m → vA/vB = 10/9. A beats C by 200 m → vA/vC = 1000/800 = 5/4. Uniform speeds. Concept / Approach: vB/vC = (vA/vC)/(vA/vB) = (5/4)/(10/9) = 9/8. In a 1350 m race for B, C covers 1350 * (8/9). Margin = difference. Step-by-Step Solution: C's distance = 1350 * 8/9 = 1200 mB's win margin = 1350 − 1200 = 150 m Verification / Alternative check: Scaling to any distance preserves the 9:8 proportion; 150 m follows directly. Why Other Options Are Wrong: 120/210 m are off the proportion; 1200 m is a misread of C’s distance, not the margin. Common Pitfalls: Confusing margin with distance covered by the slower runner. Final Answer: 150 m

Question 4

Beating distance from 1 km times: A runs 1 km in 3 min 10 s (190 s), and B runs 1 km in 3 min 20 s (200 s). Starting together for 1 km, by what distance does A beat B?

Accepted Answer

Introduction / Context: Given individual 1 km times, compute how far the slower runner has run when the faster finishes; the shortfall is the winning distance. Given Data / Assumptions: A: 1000 m in 190 s; B: 1000 m in 200 s; uniform speeds. Concept / Approach: At t = 190 s (A's finish), B's distance is (190/200)*1000. Margin is 1000 minus this distance. Step-by-Step Solution: B's distance at 190 s = 1000 * 190/200 = 950 mWinning margin = 1000 − 950 = 50 m Verification / Alternative check: Speeds are 1000/190 and 1000/200 m/s; proportional comparison yields the same 50 m difference. Why Other Options Are Wrong: 30/36/60 are typical distractors from ratio misapplication. Common Pitfalls: Subtracting times instead of comparing distances at the same time instant. Final Answer: 50 m

Question 5

Billiards handicap translation: In a 120-point game, A can give B 20 points and C 30 points. In a 90-point game, how many points can B give C?

Accepted Answer

Introduction / Context: Convert the given handicaps to speed ratios and then apply to the new total. This preserves relative strengths across game lengths. Given Data / Assumptions: A vs B: 120 vs 100 → 6:5. A vs C: 120 vs 90 → 4:3. Uniform rates. Concept / Approach: B : C = (A : C) / (A : B) = (4/3) / (6/5) = 10/9. B is faster than C in the ratio 10:9. Step-by-Step Solution: In a 90-point game for B: when B = 90, C = 90 * (9/10) = 81So B can give C = 90 − 81 = 9 points Verification / Alternative check: Alternatively, at C = 90, B = 100, which implies a 10-point edge in C's frame; translating to B's 90-point frame yields 9 as above. Why Other Options Are Wrong: 18/6/3 do not align with the 10:9 ratio at 90 total. Common Pitfalls: Using differences directly; always use ratios across different totals. Final Answer: 9 points

Question 6

Overtaking time with staggered starts: Arun starts from P at 6:00 am at 8 km/h. Bhaskar starts from P at 7:30 am at 12 km/h, both running in the same direction. At what time does Bhaskar overtake Arun?

Accepted Answer

Introduction / Context: Overtaking with different start times uses relative speed and initial lead distance acquired before the faster runner starts. Given Data / Assumptions: Arun: 8 km/h from 6:00 am. Bhaskar: 12 km/h from 7:30 am (1.5 h later). Same route and direction; constant speeds. Concept / Approach: Lead distance at 7:30 am is d = 8 * 1.5 = 12 km. Relative speed = 12 − 8 = 4 km/h. Catch time after 7:30 am is d / relative speed. Step-by-Step Solution: Lead at 7:30 am = 12 kmRelative speed = 4 km/hCatch time = 12 / 4 = 3 hOvertake time = 7:30 am + 3 h = 10:30 am Verification / Alternative check: Distances by 10:30 am: Arun = 4.5 h * 8 = 36 km; Bhaskar = 3 h * 12 = 36 km – same point. Why Other Options Are Wrong: 9:00/11:30 do not satisfy both distances equally; 1:00 am is nonsensical in this context. Common Pitfalls: Forgetting to compute Arun’s initial lead before Bhaskar starts. Final Answer: 10 : 30 am

Question 7

Catch-up times from staggered starts and different speeds: A, B, C walk 1 km in 5 min, 8 min, and 10 min, respectively. C starts at t = 0, B starts 1 min later, and A starts 2 min later than C. After A starts, when does A catch B and C (times in min…

Accepted Answer

Introduction / Context: With staggered starts, catch-up times use relative speeds and the head start each person has when A begins. We measure times from A's start instant. Given Data / Assumptions: Speeds: vA = 1/5 km/min = 0.2; vB = 1/8 = 0.125; vC = 1/10 = 0.1. Start times: C at 0, B at 1, A at 2 (minutes). Uniform speeds, straight line. Concept / Approach: Let x be minutes after A starts. Distances from the common start are equated to find catch-up times: 0.2x = 0.125(x + 1) for B; 0.2x = 0.1(x + 2) for C. Step-by-Step Solution: A vs B: 0.2x = 0.125(x + 1) ⇒ 0.2x = 0.125x + 0.125 ⇒ 0.075x = 0.125 ⇒ x = 5/3 minA vs C: 0.2x = 0.1(x + 2) ⇒ 0.2x = 0.1x + 0.2 ⇒ 0.1x = 0.2 ⇒ x = 2 min Verification / Alternative check: Convert to metres and seconds; the same proportions hold, confirming the times. Why Other Options Are Wrong: Other time pairs do not solve both linear equations simultaneously. Common Pitfalls: Measuring from absolute time instead of from A’s start; mixing minutes and hours. Final Answer: 5/3 min, 2 min

Question 8

A 10 km race is held on an 800 m circular track. Two runners P and Q compete, and their speeds are in the ratio 5 : 4. When the winner completes the full 10 km, how many complete times does the winner overtake (lap) the other on the track?

Accepted Answer

Introduction / Context: This problem tests “relative pace” on a circular track. When two runners move at constant but different speeds, each full lap gained by the faster runner over the slower runner corresponds to the faster accumulating an extra one-track-length of distance relative to the slower. The question asks for how many complete overtakings (full laps gained) occur by the time the winner finishes the race distance. Given Data / Assumptions: Total race distance for the winner = 10 km = 10,000 m. Track length (circumference) = 800 m. Speed ratio P:Q = 5:4 (constant speeds; no stops). Concept / Approach: The slower runner’s covered distance at the instant the faster finishes is proportional to the speed ratio. If the winner covers 10,000 m, the other covers 10,000 * (slower_speed / faster_speed). The number of complete overtakings equals (distance lead) / (track length), counting only full laps (integer part). Step-by-Step Solution: Let P be the faster. When P runs 10,000 m, Q runs 10,000 * (4/5) = 8,000 m.Lead gained by P over Q = 10,000 − 8,000 = 2,000 m.Each complete overtake corresponds to 800 m of relative lead.Complete overtakings = 2,000 / 800 = 2.5 ⇒ full overtakings = 2 times. Verification / Alternative check: If you simulate by laps: P’s laps = 10,000/800 = 12.5; Q’s laps = 8,000/800 = 10. The difference in completed laps is 2, confirming two full passes. Why Other Options Are Wrong: 1 time or 3–4 times do not match the 2,000 m relative lead; only two full 800 m gains fit before the finish. Common Pitfalls: Counting the fractional (half) lap as a full overtake or mistakenly computing (5−4)/4 laps instead of using total lead divided by track length. Final Answer: 2 times

Question 9

A runs at 1 2/3 (that is, 5/3) times the speed of B. If A gives B a start of 40 m and they finish together, how far from A’s starting line should the winning post be placed?

Accepted Answer

Introduction / Context: Starts (head-starts) in races shift the effective distance each runner must cover. If two runners finish together, the time taken by both is equal. With a start advantage to the slower runner, we relate distances to speeds using the time equality for the same finishing instant. Given Data / Assumptions: A’s speed : B’s speed = 5 : 3. Head-start to B = 40 m. Let race length (distance A runs) be D. Concept / Approach: Time = distance / speed. With equal finish time: D / vA = (D − 40) / vB. Substitute vA / vB = 5/3 to solve for D. Step-by-Step Solution: D / (5/3 vB) = (D − 40) / vB ⇒ (3D/5) = D − 40.Multiply by 5: 3D = 5D − 200 ⇒ 2D = 200 ⇒ D = 100 m. Verification / Alternative check: Times: A takes 100 / (5k) = 20/k; B takes 60 / (3k) = 20/k. Equal ⇒ consistent. Why Other Options Are Wrong: 75, 125, or 200 m do not satisfy the equality when substituted. Common Pitfalls: Interpreting “start” as extra distance for A instead of reduced distance for B, or inverting the given speed ratio. Final Answer: 100 m

Question 10

In a 20 m race, A runs at 2 m/s. A gives B a 10 m head-start and still beats B by 5 seconds. What is B’s speed (in m/s)?

Accepted Answer

Introduction / Context: Head-starts alter distances while speeds remain constant. If one runner finishes earlier by a known time margin, we set up times for each runner and equate the difference to the given head-to-head gap in seconds. Given Data / Assumptions: Total race distance from A’s mark = 20 m. A’s speed = 2 m/s ⇒ A’s time = 20 / 2 = 10 s. B starts 10 m ahead, so B needs to cover 10 m to finish. A beats B by 5 s ⇒ B’s time = 10 + 5 = 15 s. Concept / Approach: B’s average speed over their race distance equals distance / time. With B’s distance 10 m and time 15 s, compute vB. Ensure dimensions match. Step-by-Step Solution: tA = 20/2 = 10 s.tB = tA + 5 = 15 s.vB = distance / time = 10 / 15 = 2/3 m/s. Verification / Alternative check: At 2/3 m/s, B covers 10 m in 15 s; A indeed finishes at 10 s, leading by 5 s, as stated. Why Other Options Are Wrong: 1 m/s or 4/3 m/s would produce smaller/larger finish gaps than 5 s; 2 m/s is A’s speed, not B’s. Common Pitfalls: Using 20 m for B’s distance (ignoring the 10 m start) or mixing up “beats by 5 s” with “beats by 5 m”. Final Answer: 2/3 m/s

Question 11

Shantanu cycles along the boundary of a square field ABCD starting at corner A. After 1/2 hour he reaches corner C (the corner diagonally opposite A) by following the boundary. If his cycling speed is 16 km/h, what is the area of the square field?

Accepted Answer

Introduction / Context: On a square, traveling from one corner to the diagonally opposite corner along the boundary equals half the perimeter. If the traveler’s speed and the time are known, the boundary distance is known, from which we find the side and thus the area. Given Data / Assumptions: Time = 1/2 h. Speed = 16 km/h ⇒ distance covered = 16 * 1/2 = 8 km. Boundary distance from A to opposite corner C = half the perimeter = 2s, where s is the side of the square. Concept / Approach: Since 2s = 8 km ⇒ s = 4 km. Area of square = s^2 = 16 sq km. Compare to options. Step-by-Step Solution: Distance along boundary to opposite corner = 2s.Given distance = 8 km ⇒ 2s = 8 ⇒ s = 4 km.Area = s^2 = 16 sq km. Verification / Alternative check: Perimeter would be 4s = 16 km; half is 8 km, exactly what he traveled in 1/2 h at 16 km/h. Why Other Options Are Wrong: 8, 9, or 19 sq km do not match the computed 16 sq km. Hence “None of the above” is correct. Common Pitfalls: Confusing boundary travel with the straight diagonal (which is s√2, not 2s) or misusing time–speed–distance. Final Answer: None of the above (actual area = 16 sq km)

Question 12

In an 800 m race, A can beat B by 40 m (so B covers 760 m when A finishes 800 m). In a 500 m race, B can beat C by 5 m (so C covers 495 m when B finishes 500 m). By how much does A beat C in a 200 m race?

Accepted Answer

Introduction / Context: Racing comparison problems are solved by converting finish advantages into speed ratios, then chaining those ratios. Once the overall speed ratio between two athletes is known, scale it to any race length to compute finish separations. Given Data / Assumptions: A vs B (800 m): vA/vB = 800/760 = 20/19. B vs C (500 m): vB/vC = 500/495 = 100/99. We need A vs C over 200 m. Concept / Approach: Multiply ratios: vA/vC = (vA/vB) * (vB/vC) = (20/19) * (100/99) = 2000/1881. When A runs 200 m, C runs 200 / (vA/vC). The difference gives the margin by which A beats C. Step-by-Step Solution: vA/vC = 2000/1881 ≈ 1.0633.C’s distance when A finishes 200 m = 200 * 1881 / 2000 = 188.1 m.Lead = 200 − 188.1 = 11.9 m. Verification / Alternative check: Using proportional times gives the same outcome because distance at equal time is proportional to speed. Why Other Options Are Wrong: 12.7 m and the centimeter-scaled distractors are not consistent with the compounded ratio 2000/1881. Common Pitfalls: Adding raw meter advantages directly across different race lengths; advantages must be translated to speed ratios first. Final Answer: 11.9 m

Question 13

Three runners Ram, Karan, and Mohan start from a fixed post around an oval at speeds 5 km/h, 5 1/7 km/h, and 5 1/3 km/h respectively. They all return to the fixed post exactly in 3 3/4 hours. What is the greatest possible track length (in km) consis…

Accepted Answer

Introduction / Context: When multiple runners all return to a common starting post at the same time, the track length must divide the distance each runner covers in that time, so that each completes an integer number of laps. The largest feasible track length equals the greatest common divisor (GCD) of the three distances covered. Given Data / Assumptions: Time t = 3 3/4 h = 15/4 h. Speeds: v1 = 5, v2 = 5 1/7 = 36/7, v3 = 5 1/3 = 16/3 (km/h). Concept / Approach: Distances: d1 = v1*t = 5*(15/4) = 75/4; d2 = (36/7)*(15/4) = 135/7; d3 = (16/3)*(15/4) = 20. The maximum track length L is the GCD of {d1, d2, d3}. Step-by-Step Solution: Express with a common denominator 28: d1 = 525/28, d2 = 540/28, d3 = 560/28.GCD(525, 540, 560) = 5; therefore L = 5/28 km. Verification / Alternative check: Check laps: (525/28)/(5/28) = 105; (540/28)/(5/28) = 108; (560/28)/(5/28) = 112. All integers. Why Other Options Are Wrong: Other fractions do not divide each distance into an integer number of laps. Common Pitfalls: Using LCM of times without converting to distances, or taking an average instead of the GCD. Final Answer: 5 / 28 km

Question 14

A and B start together and run in the same direction on a 3 km circular track. Their speeds are 200 m/min and 150 m/min respectively. After how long will they both be together again at the starting point?

Accepted Answer

Introduction / Context: To be together again at the start, each runner must have completed an integer number of laps. Thus, the rendezvous time at the start equals the least common multiple (LCM) of their individual lap times. Given Data / Assumptions: Track length = 3,000 m. Speeds: 200 m/min and 150 m/min. Concept / Approach: Lap times: T1 = 3,000/200 = 15 min; T2 = 3,000/150 = 20 min. Meeting at the start occurs at LCM(15, 20) min. Step-by-Step Solution: LCM(15, 20) = 60 min = 1 hour. Verification / Alternative check: After 60 min, runner A has done 4 laps; runner B has done 3 laps—both at the start again. Why Other Options Are Wrong: Other times are not common multiples of both 15 and 20 minutes. Common Pitfalls: Using relative speed to compute the time they meet anywhere on track; here the location “starting point” forces an LCM approach. Final Answer: 1 hour

Question 15

A can beat B by 5 m in a 100 m race (so B covers 95 m when A covers 100 m). B can beat C by 15 m in a 200 m race (so C covers 185 m when B covers 200 m). By how many meters will A beat C in a 400 m race?

Accepted Answer

Introduction / Context: Convert finish advantages to speed ratios and chain them to compare two runners who have not raced directly. Once vA/vC is obtained, scale to the target race to compute the margin. Given Data / Assumptions: vA/vB = 100/95 = 20/19. vB/vC = 200/185 = 40/37. Target race length for A = 400 m. Concept / Approach: vA/vC = (20/19)*(40/37) = 800/703. When A runs 400 m, C runs 400 / (800/703) = 703/2 = 351.5 m, giving the margin. Step-by-Step Solution: vA/vC = 800/703 ≈ 1.138.C’s distance at A’s finish = 351.5 m.Lead = 400 − 351.5 = 48.5 m. Verification / Alternative check: Time ratio cross-check gives the same margin because distance at equal time ∝ speed. Why Other Options Are Wrong: 50.5, 45.5, 42.5 m do not align with the compounded ratio. Common Pitfalls: Adding meter advantages directly rather than converting to speed ratios first. Final Answer: 48.5 m

Question 16

In billiards, A can give B 6 points in a game to 50, and A can give C 13 points in a game to 65. In a game to 55, how many points can B give C (so that the game is fair)?

Accepted Answer

Introduction / Context: Handicap statements like “A gives B 6 in 50” imply proportional scoring ability: when A scores 50, B scores 44. Converting these into ability ratios allows fair handicaps to be computed for new target totals. Given Data / Assumptions: A vs B to 50: B scores 44 when A scores 50 ⇒ A:B = 50:44 = 25:22. A vs C to 65: C scores 52 when A scores 65 ⇒ A:C = 65:52 = 5:4. Concept / Approach: From A:B = 25:22 and A:C = 5:4 ⇒ take A = 25 units; then B = 22, C = (4/5)×25 = 20. Hence B:C = 22:20 = 11:10. In a fair game to 55 for B, C should be proportionally at 55×(10/11) = 50. Step-by-Step Solution: Fair target mapping: When B scores 55, C would score 50.Therefore B can give C = 55 − 50 = 5 points. Verification / Alternative check: Scaling any common multiple maintains the 11:10 ratio. Why Other Options Are Wrong: 7, 8, or 10 would distort the B:C proportionality implied by the original handicaps. Common Pitfalls: Adding or averaging handicaps directly without converting to ability ratios. Final Answer: 5

Question 17

In a 200 m race, B can give A a start of 10 m (so A runs 190 m when B runs 200 m). In the same 200 m race, C can give B a start of 20 m (so B runs 180 m when C runs 200 m). How much start can C give A in the same race?

Accepted Answer

Introduction / Context: Head-starts yield speed ratios. Chain them to compare two competitors who have not raced directly. Then, compute the distance A would cover when C finishes the 200 m, and infer the start C can give A. Given Data / Assumptions: B vs A: vB/vA = 200/190 = 20/19. C vs B: vC/vB = 200/180 = 10/9. Target race = 200 m with C finishing the full 200 m. Concept / Approach: vC/vA = (vC/vB)*(vB/vA) = (10/9)*(20/19) = 200/171. When C runs 200 m, A covers 200*(vA/vC) = 200*(171/200) = 171 m, so the start equals 200 − 171. Step-by-Step Solution: A’s distance at C’s finish = 171 m.Hence C can give A = 200 − 171 = 29 m. Verification / Alternative check: Time equality check confirms the same value since distance is proportional to speed at a common finish time. Why Other Options Are Wrong: 30, 27, 25 m do not match the chained ratio result 200/171. Common Pitfalls: Adding start values directly (10 + 20) instead of chaining speed ratios, which is incorrect. Final Answer: 29 m

Question 18

A runs 500 m in 30 s; B runs 500 m in 35 s. In a 1 km race, how much head-start (in meters) should A give B so that they finish in a dead heat?

Accepted Answer

Introduction / Context: For a dead-heat with a head-start, the slower runner’s race distance is shorter, and the finish times must be equal. Compute each runner’s speed from the given 500 m times and then set equal times for the 1 km event to solve the required start. Given Data / Assumptions: vA = 500/30 = 50/3 m/s. vB = 500/35 = 100/7 m/s. Let head-start to B be x meters. A runs 1,000 m; B runs (1,000 − x) m. Concept / Approach: Equal finish time: 1,000 / vA = (1,000 − x) / vB. Solve for x. Convert to a mixed fraction as needed. Step-by-Step Solution: 1,000 / (50/3) = (1,000 − x) / (100/7).Left: 60 s. Right: (1,000 − x)*7/100.Equate: 60 = (7,000 − 7x)/100 ⇒ 6,000 = 7,000 − 7x ⇒ 7x = 1,000 ⇒ x = 1,000/7 = 142 6/7 m. Verification / Alternative check: Plugging x back equates both times to 60 seconds. Why Other Options Are Wrong: Nearby fractions give unequal finish times; only 1,000/7 m equalizes them. Common Pitfalls: Using average speeds or subtracting times directly without adjusting distances. Final Answer: 142 6/7 m

Question 19

A’s speed is 1 3/8 times B’s speed (that is, 11/8 of B). In a race, A gives B a start of 300 m and they finish together. What should be the total race length (distance A runs) so both reach the winning post simultaneously?

Accepted Answer

Introduction / Context: When a faster runner gives a head-start but still aims to finish together, the time equality relates the full distance for the faster to the reduced distance for the slower via their speed ratio. Given Data / Assumptions: vA = (11/8) vB. B’s head-start = 300 m. Let race length (A’s distance) be L; B runs L − 300. Concept / Approach: Equal finish times: L / vA = (L − 300) / vB. Substitute vA and solve for L. Step-by-Step Solution: L / ((11/8) vB) = (L − 300)/vB ⇒ (8L/11) = L − 300.Multiply by 11: 8L = 11L − 3,300 ⇒ 3L = 3,300 ⇒ L = 1,100 m. Verification / Alternative check: Times become equal: A’s time = 1,100 / (11/8 vB) = 800 / vB; B’s time = 800 / vB. Why Other Options Are Wrong: Other lengths do not satisfy the derived equality. Common Pitfalls: Subtracting 300 from the faster runner’s distance, or inverting the 11/8 ratio. Final Answer: 1,100 m

Question 20

In a 1 km race, A beats B by 80 m (i.e., B covers 920 m when A covers 1,000 m), and B beats C by 60 m (i.e., C covers 940 m when B covers 1,000 m). By how many meters does A beat C in the same 1 km race?

Accepted Answer

Introduction / Context: Use chained speed ratios to compare two runners not racing directly. Once vA/vC is known, compute C’s distance at the instant A completes 1,000 m and subtract from 1,000 m to obtain the margin. Given Data / Assumptions: vA/vB = 1000/920 = 25/23. vB/vC = 1000/940 = 50/47. Concept / Approach: vA/vC = (25/23)*(50/47) = 1250/1081. When A finishes 1,000 m, C has covered 1,000 * 1081/1250 = 864.8 m. Step-by-Step Solution: Margin = 1,000 − 864.8 = 135.2 m. Verification / Alternative check: Proportional-time check matches the computed distance for C. Why Other Options Are Wrong: Other values do not match the compounded ratio 1250/1081. Common Pitfalls: Adding meter advantages directly; this is invalid across different base races. Final Answer: 135.2 m

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